CS 151 Complexity Theory Spring Final Solutions. L i NL i NC 2i P.

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1 CS 151 Complexity Theory Spring 2017 Posted: June 9 Final Solutions Chris Umans 1. (a) The procedure that traverses a fan-in 2 depth O(log i n) circuit and outputs a formula runs in L i this can be done by a recursive depth-first traversal, which only requires 1 bit of information ( left or right ) at each level of recursion. The procedure for FVAL (Lecture 2) runs in log-space, so on a formula of size 2 O(logi n), it runs in O(log i n) space. Using space-efficient composition of the logspace procedure that generates the circuit together with these two procedures we obtain a procedure to evaluate an NC i circuit on a given input in only O(log i n) space, as required. (b) The configuration graph for an NL i machine on input x of length n has at most 2 O(logi n) nodes. The input x is accepted if and only if there is a path from the start node s to the accept node t in this graph. We can construct the incidence matrix A of this graph (with ones on the diagonal), and we observe that A = A 2m, for m = O(log i n) has a one in position s, t if and only if there is a path of length at most 2 m from s to t (here we are using Boolean matrix multiplication). We can square matrix A with a O(log A ) = O(log i n) depth circuit. We repeat this squaring m times, to compute A. The repeated squaring entails m sequential copies of the squaring circuit, which has depth O(log i n). The total depth is O(log 2i n). (c) Suppose we show NL i NC 2i for some i > 1. Then we have L i NL i NC 2i P. However, we know by the Space Hierarchy Theorem that L is strictly contained in L i for i > 1. Thus we would have proved L P. In fact, we would have proved something stronger: that NC 1 NC 2, since an equality would collapse all of the hierarchy to NC 1, including NC 2i (and then we would have NC 1 = L = L i = NL i = NC 2i, contradicting the Space Hierarchy Theorem). 2. (a) Fix an x {0, 1} n and a y {0, 1} k. Imagine that we have already chosen M. In order to have h M,b (x) = y, we must have Mx + b = y or equivalently y Mx = b. This happens with probability exactly 2 k since b is chosen uniformly from {0, 1} k. For the second part, we know that x 1 x 2. Thus there must be a position i in which they differ. WLOG, assume (x 1 ) i = 1 and (x 2 ) i = 0. Imagine that we have already chosen all of M except for the i-th column, and denote by M the matrix M with 0s in the i-th column. Let us denote by a {0, 1} k our choice of the i-th column of M. Note that h M,b (x 1 ) = M x 1 + a + b and h M,b (x 2 ) = M x 2 + b. Thus we are interested in the probability that a + b = y 1 M x 1 and b = y 2 M x 2. Since each of a and b are chosen uniformly and independently from {0, 1} k, this happens with probability 2 2k. More precisely, there is a 2 k chance of choosing b equal to the fixed vector y 2 M x 2, and then given the choice of b, there is a 2 k chance of choosing a equal to y 1 M x 1 b. 0-1

2 0-2 (b) Here is a 2-round AM protocol for largeset. The common input is (C, k). Arthur picks a random y and a random k n matrix M and b {0, 1} k as above and sends them to Merlin. Merlin replies with an x {0, 1} n. Arthur accepts iff C(x) = 1 and h M,b (x) = y. We have to show completeness and soundness for this protocol. For completeness, set A = {x : C(x) = 1}, and observe that if A 3(2 k ) then the inequality from the problem statement gives us: Pr [ x A h M,b(x) = y] 1 2k M,b,y A 2 3. Thus given a YES instance, with probability at least 2/3, Merlin has a reply that will cause Arthur to accept. For soundness, again set A = {x : C(x) = 1}, and observe that if A (2 k )/3 then from part (a), we have that for each fixed x A, Pr [h M,b(x) = y] = 2 k. M,b,y Taking a union bound over all x A, we get Pr [ x A h M,b(x) = y] 2 k A 1 M,b,y 3. Thus given a NO instance, the probability that Merlin has a reply that will cause Arthur to accept is at most 1/3. Finally, apply the transformation from Problem Set 6, Problem 3, to this protocol to achieve perfect completeness. 3. Let L be a language in PSPACE, and let x be an input of length n. Using the given fact, together with the assumption that PSPACE has polynomial-size circuits, there is a polynomial size circuit C that computes the (honest) prover s messages as a function of x and the messages seen so far, in the IP protocol for L. We need to describe a MA protocol for L. We have Merlin send the circuit C in the first round. Then Arthur simulates the IP protocol for L with input x, evaluating C to determine the prover s messages at each step. This entails flipping polynomially many coins, and evaluating the circuit C polynomially many times. In the end Arthur accepts if the Verifier he is simulating would have accepted. Now, if x is in L, then there exists a Merlin message that will cause Arthur to accept with probability at least 2/3 namely, the circuit that correctly computes the Prover messages in the IP protocol for L. On the other hand, if x L, then no matter what C is sent in the first round, Arthur will reject with probability at least 2/3, because of the soundness guarantee for the IP protocol. I.e., the evaluations of any circuit C correspond to some (possibly dishonest) prover, and we know that when x L, no prover can cause the Verifier to accept with more than 1/3 probability. This shows that PSPACE MA. We know that MA PSPACE unconditionally, so we conclude that under the assumption PSPACE P/poly, we have PSPACE = MA.

3 (a) For a language L S p 2, we have Thus L Σ p 2. We also have: x L y z (x, y, z) R x L z y (x, y, z) R y z (x, y, z) R x L y z (x, y, z) R z y (x, y, z) R x L z y (x, y, z) R and so L Π p 2. We conclude that Sp 2 (Σp 2 Πp 2 ). (b) Let L be an arbitrary language in P NP and let M be an oracle Turing Machine that decides L in time n c for some constant c. Fix an input x. Without loss of generality we standardize M so that its oracle is SAT, and all of its oracle queries are 3-CNF formulas with m variables. We describe the behavior of two machines M 1 and M 2 that run in polnyomial time; these are then converted into the circuits C 1 and C 2 that the reduction produces from x. Machine M 1 simulates machine M on input x, until M makes an oracle query: φ SAT?. At this point M 1 consults its input y, and reads m + 1 bits of y. If the first bit is 0, it checks if the remaining m bits are a satisfying assignment to φ; if they are it continues simulating M as if M had received a yes answer to its query, otherwise it rejects. If the first bit is 1, it discards the remaining m bits, and continues simulating M as if M had received a no answer to its query. We continue in this fashion, reading successive (m 1)-bit segments of y as (our simulation of) M encounters successive oracle queries. We stop when M 1 has simulated x c steps of M, at which point it accepts. Machine M 2 does exactly the same thing as M 1, except that it accepts at the end iff M would have accepted at this point. Note that depending on y, this may or may not agree with what M SAT actually does on input x. However, we claim that the lexicographically first y that M 1 accepts causes M 2 to correctly simulate M SAT on input x. This is true because at each query φ, the lexicographically first m + 1 bits that will cause M 1 to continue its simulation are either (1) 0 followed by the lexicographically first satisfying assignment to φ if φ SAT, or (2) 1 followed by all zeros if φ SAT. In case (1) our simulation proceeds as if it received a yes answer to the query and in case (2) our simulation proceeds as if it received a no answer; in both cases this correctly simulates M SAT. We conclude that M 2 accepts the lexicographically first y accepted by M 1 iff M SAT accepts x, as required. We also should argue that the problem is in P NP, but this is easy, because we can do a binary search (using the NP oracle) to identify the lexicographically first y accepted by C 1, and then plug it into C 2. (c) We argue that lex-first-acceptance is in S p 2. Let (C 1, C 2 ) be an instance of lexfirst-acceptance. Define the function f(y, y ) to be C 2 (y min ) where y min is the lexicographically first among y, y that C 1 accepts; or 0 if C 1 (y) = C 1 (y ) = 0. We claim that (C 1, C 2 ) lex-first-acceptance y y f(y, y ) = 1 (C 1, C 2 ) lex-first-acceptance y y f(y, y ) = 0

4 0-4 This is easily seen by taking y to be the lexicographically first string accepted by C 1 in the first case, and y to be the lexicographically first string accepted by C 1 in the second case. Since lex-first-acceptance is P NP -complete, we conclude that P NP S p 2. (d) By error reduction, we may assume that for every language L in MA there is a language R in P for which We claim that x L y Pr z [(x, y, z) R] = 1 x L y Pr z [(x, y, z) R] < 2 y. x L y z (x, y, z) R x L z y (x, y, z) R, which implies that L S P 2 as required. The first part is obvious from the definitions. For the second part, observe that implies (by the union bound) This implies z y (x, y, z) R as required. y Pr z [(x, y, z) R] < 2 y Pr z [ y (x, y, z) R] < 2 y 2 y = 1. (0.1) (e) Given a language L BPP, we can use strong error reduction to produce a probabilistic polynomial time TM M for which: x L Pr y [M(x, y) = 1] 1 2 y 1/3 2 y x L Pr y [M(x, y) = 0] 1 2 y 1/3 2 y. We split y into two equal-length substrings y = u v. Our predicate R is simply R(x, u, v) = M(x, u v). Now, if x L, then it must be that u v R(x, u, v) = 1, for if not, then u v R(x, u, v) = 0 which implies that M(x, y) = 0 for at least 2 y /2 2 y 1/3 values of y, a contradiction. Similarly, if x L, then it must be that v u R(x, u, v) = 0, for if not, then v u R(x, u, v) = 1 which implies that M(x, y) = 1 for at least 2 y /2 2 y 1/3 values of y, a contradiction. We conclude that L S p 2 and therefore BPP Sp 2 as required. Another solution is to observe that BPP is contained in (2-sided error) MA and apply the previous part! (f) The following notation will be useful: given a circuit C with a single Boolean output, let C be the circuit derived from C that uses C as if it were a circuit for SAT to actually find

5 0-5 a satisfying assignment (via the self-reducibility of SAT). If at any point in the repeated applications of C, there is an inconsistent answer, C outputs some fixed string, say, the all-zeros string. So, C has as many outputs as inputs, and C poly( C ), and if C is a circuit correctly computing SAT, then C will correctly output a satisfying assignment if there is one. Let L be a language in Π p 2, so we have x L y z (x, y, z) R x L y z (x, y, z) R for some language R P. Observe that the language L = {(x, y) : z (x, y, z) R} is in NP, and so given a pair (x, y) we can use a procedure that solves SAT and actually returns a satisfying assignment if there is one to find z for which (x, y, z) R if such a z exists. Define R to be the language consisting of exactly the triples (x, C, y) for which using C, we obtain a z for which (x, y, z) R. Notice that R can be evaluated in polynomial time. We are assuming that SAT has polynomial-size circuits. If x L, then there exists a circuit C (the one that computes SAT) for which for all y, C will successfully find a z that causes R to accept. Thus x L C y (x, C, y) R. If x L, then there is some y for which z (x, y, z) R. Thus for all C, (x, C, y ) R, because no matter what z we find using C, it will not be the case that (x, y, z) R. Therefore x L y C (x, C, y) R. We conclude that L S p 2. We have shown that Π p 2 Sp 2. Since Sp 2 is closed under complement, we also have that Σ p 2 Sp 2. Using part (a), we have Πp 2 = Σp 2 = Sp 2, and so the PH collapses to Σp 2 = Sp 2 as required.

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